NOTE: There is most likely a typo in the problem since none of the dimension analysis is adequate for the calculation of the EMF. We believe that the correct answer is B and that there is a type in the expression for Teslas.

In order to solve this problem let’s write the EMF equation which is given as

$$
EMF=BLv
$$

The dimension analysis of each term gives that

$$
B=textrm{T}=frac{textrm{N}}{textrm{m}cdottextrm{A}}
$$

$$
L=textrm{m}
$$

$$
v=frac{textrm{m}}{textrm{s}}
$$

After we insert this into the EMF equation we get

$$
textrm{V}=frac{textrm{N}}{textrm{m}cdottextrm{A}}cdot mcdotfrac{textrm{m}}{textrm{s}}
$$

So the correct answer should be B, if we assume that there is a typo in the problem.

In order to solve this problem, we are going to use the EMF equation which gives that

$$
EMF=BLv
$$

From here, one can express the magnetic field as

$$
B=frac{EMF}{Lv}=frac{4.2}{0.427times 0.186}
$$

Finally, one obtains that

$$
boxed{B=52.9textrm{ T}}
$$

So the correct answer is D.

In order to solve this problem, we are going to use the EMF equation which gives that

$$
EMF=BLv
$$

Which after we plug in the values gives

$$
EMF=1.4times 0.15times 0.12
$$

Finally, we have that

$$
boxed{EMF=0.025textrm{ V}}
$$

So the correct answer is C.

In order to solve this problem, we are going to use the transformer equation which is given as
$frac{I_s}{I_p}=frac{V_p}{V_s}=frac{N_p}{N_s}$
Now, we can express the current in the primary circuit

$$
I_p=frac{V_s}{V_p}I_s=frac{13}{91}times 1.9
$$

Which gives that

$$
boxed{I_p=0.27textrm{ A}}
$$

So the correct answer is A.

In order to solve this problem we are going to use the formulas for the effective voltage, effective current and Ohm’s law. So, let’s do it.

We start from the formula for the effective voltage which says

$$
V_{eff}=frac{sqrt{2}}{2}V_{max}
$$

So after we plug the values

$$
V_{eff}=frac{sqrt{2}}{2}202
$$

$V_{eff}=142.8textrm{V}$
Now, we can use Ohm’s law to find the effective current

$$
I_{eff}=frac{V_{eff}}{R}=frac{142.8}{480}
$$

Finally

$$
boxed{I_{eff}=0.298textrm{A}}
$$

So the correct answer is A.

In order to solve this problem we are going to use the power law which says that the current in the first case will be given as

$$
I_1=frac{P}{V_1}=frac{800}{160}=5textrm{ A}
$$

Now, the dissipated power is given as

$$
P_1=RI_1^2=2times 5^2=50textrm{W}
$$

In the second case we have that

$$
I_2=frac{P}{V_2}=frac{800}{960}=0.833textrm{ A}
$$

and the dissipated power is given as

$$
P_2=RI_2^2=2times 0.83^2=1.4textrm{W}
$$

We conclude that higher voltages yield with lower power loss which explains the high-voltage power lines connecting large distances.